What is the balanced reaction when silver nitrate reacts with magnesium chloride to produce silver chloride and magnesium nitrate?

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Stoichiometry and Balancing Reactions/Chemical Equations

Stoichiometry is founded based on the law of conservation of mass. It involves the using of relationships between reactants and/or products in a chemical reaction to determine the desired quantitative data. It comes from the Greek word"stoikhein" that means element and greek word "metron" that means measure, so stoichiometry means the measure of elements.

* In order to use stoichiometry and to run calculations about chemical reactions. It is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding in how to balanced reactions.

The study of quantities of products and their relationship to quantities of reactants in a chemical reaction is called stoichiometry. A balanced chemical equation is not just a qualitative description of a reaction, it is also quantitative. It tells us not just what the reactants and products are, but also the relative amounts of reactants and products that are involved in the reaction. The coefficients in the balanced equation tell us the relative number of formula units for each compound that participate in or are produced by the reaction. The following equation can be read as one formula unit of magnesium chloride reacts with two formula units of silver nitrate to give two formula units of silver chloride and one formula unit of magnesium nitrate.

However many formula units of magnesium chloride we start with, the equation tells us that we will need two times that number of formula units of silver nitrate, and we will get as a product twice as many formula units of silver chloride as formula units of magnesium chloride that we started with. It is convenient to choose Avogadro's number as the number of formula units we consider, because/then we can read the equation in terms of moles of the compounds involved: one mole of magnesium chloride reacts with two moles of silver nitrate to give two moles of silver chloride and one mole of magnesium nitrate.'

We relate the amounts of two compounds in the reaction using a mole ratio derived from the balanced chemical equation. For example, to calculate the amount of silver chloride produced when7.35 mol of magnesium chloride reacts with silver nitrate, we use the ratio 2 mol AgCl per 1 mol MgCl2.

*Mole Ratio/Molar Ratio

-is the relationship of the numerical coefficient of the reactants to the product
Let's use mole ratios to answer the following question. If 0.276 mol of silver chloride is produced in a reaction of magnesium chloride with silver nitrate, what amount of each reactant has been used?To solve the problem we use the appropriate mole ratios from the equation to 'convert' the amount of product formed to the amount of each reactant that must have been used:Notice how the units 'mol AgCl' cancel out so that we can be sure we have used the mole ratios the right way up to calculate that 0.138 mol of MgCl2 must have reacted with 0.276 mol of AgNO3.

Mass Relationships

In practice, quantities of reactants and products are measured by mass (or volume). Since the balanced chemical equation gives mole relationships, we must use the molar mass of each compound to convert mass in grams to moles, or moles to grams, as appropriate.For example, how much silver chloride could be produced by the complete reaction of 19.7 g of magnesium chloride with silver nitrate? In planning to solve this problem, we recognize we need to convert the mass of MgCl2 to moles using the molar mass (95.21 g MgCl2/mol MgCl2), then use the mole relationship from the equation to find the amount (moles) of silver chloride produced, then use the molar mass of silver chloride (143.3 g AgCl/mol AgCl) to convert this to a quantity in grams:This type of problem, requiring conversions from '{measured quantity of compound A} to {amount in moles of compound A} to {amount in moles of compound B} to {measured quantity of compound B}' is encountered frequently in the study of chemical reactions. Look carefully at the units in the conversion factors in the solution above to see how they relate to this strategy.For practice, let's ask two more questions about this reaction. How much silver nitrate has reacted with our 19.7 g of magnesium chloride, and how much magnesium nitrate has been produced? These questions are answered using the same strategy outlined above: {mass MgCl2} to {moles MgCl2} to {moles of compound B (using a mole ratio from the equation)} to {mass of compound B (using the molar mass of compound B)}:Looking carefully at these quantities, we see that 90.0 g of products (59.3 g of AgCl + 30.7 g of Mg(NO3)2) has been formed from 90.0 g of reactants (19.3 g of MgCl2 + 70.3 g of AgNO3). The law of conservation of mass is satisfied as we would expect, because a balanced chemical equation has the same number of each type of atom on both sides..

Limiting Reagent

In the previous example we saw that if we poured a solution containing 70.3 g of silver nitrate into a solution containing 19.7 g of magnesium chloride we would expect to get 59.3 g of silver chloride precipitated. Now suppose we pour the same silver nitrate solution into a solution containing 30.0 g of magnesium chloride. Would we get any more silver chloride precipitated? No! 70.3 g of silver nitrate is just enough to react with 19.7 g of magnesium chloride, and when that amount has reacted then all the silver nitrate is used up. The extra magnesium chloride (30.0 g - 19.7 g = 10.3 g MgCl2) is left over and remains in solution. We say that magnesium chloride is an excess reagent/reactant, and silver nitrate (the reactant which is all used) is the limiting reagent/reactant.

The amount of product formed is determined by the amount of the limiting reagent. When this is all used then no more product can be formed. When you are given quantities of two reactants, you must determine which one is the limiting reagent and use this one to calculate the quantity of product that can be formed.Putting all of these ideas together we can solve a problem of the following type:How much barium sulfate can be produced if a solution containing 10.0 g of barium acetate is mixed with a solution containing 10.0 g of potassium sulfate?First we write an equation for the reaction using the correct formulas for each compound involved:Next, we choose one of the reactants and calculate how much of the other reactant would be required to react completely with the amount we have. Let's see how much potassium sulfate (molar mass 174.3 g/mol)would react with the 10.0 g of barium acetate (molar mass 255.4 g/mol) we have to start with:Now we compare this calculated amount with the amount of potassium sulfate we actually have - 10.0 g. We see that we have more potassium sulfate than we need, so potassium sulfate is in excess and barium acetate is the limiting reagent. Finally we calculate the amount of barium sulfate (molar mass 233.4 g/mol) produced from the amount of limiting reagent that we started with.